I am interested in the value of

$$\int_0^\infty e^{-\alpha t}\frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}}\, dt $$

this is the laplace transform of the Heat kernel (changing the time variable)

This integral came up in this exercise:

Exercise 3.7. Show that the resolvent for Brownian motion is given by $$U(\alpha)f(x)=\frac{1}{\sqrt{2\alpha}}\int_{-\infty}^\infty f(y)e^{-\sqrt{2\alpha}|x-y|}dy.$$


$$U(\alpha)f=\int_0^\infty e^{-\alpha t}T(t)f\,dt,\quad\alpha>0,\tag{3.6}$$

First setps:

\begin{align} U(\alpha)f(x) &= \int_0^\infty e^{-\alpha t} T(t) f (x)\, dt \\ &= \int_0^\infty e^{-\alpha t} \int_{-\infty}^\infty \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} f (y)\, dy\, dt \\ &= \int_{-\infty}^\infty f(y)\int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt \, dy \\ \end{align}

So one can guess that

$$ \int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt = \frac{\exp{- \sqrt{2\alpha}|x-y|}}{\sqrt{2 \alpha}}$$

But How do we get such result?

Here is my attempt at proving this:

$$\int_0^\infty e^{-\alpha t} \frac{e^{-\frac{|x-y|^2}{2t}}}{\sqrt{2\pi t}} \, dt = \int_0^\infty \exp\bigg\{-\alpha t - \frac{|x-y|^2}{2t}\bigg\} \frac{1}{\sqrt{2\pi t}} \, dt \\ = \int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 + \sqrt{2\alpha}|x-y|\bigg\} \frac{1}{\sqrt{2\pi t}} \, dt\\ = \exp\bigg\{- \sqrt{2\alpha}|x-y|\bigg \}\int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 \bigg\} \frac{1}{\sqrt{2\pi t}} \, dt$$

So now I am at the point that I need to prove

$$\int_0^\infty \exp\bigg\{-\bigg(\sqrt{\alpha t} - \frac{|x-y|}{\sqrt{2t}}\bigg)^2 \bigg\} \frac{1}{\sqrt{2\pi t}} \, dt = \frac{1}{\sqrt{2 \alpha}} $$

But I don't see how to do it

  • 1
    $\begingroup$ Hint: $$ \int_0^{\infty} f\left(\left(ax-\frac{b}{x}\right)^2\right)dx=a^{-1} \int_0^{\infty} f(y^2)dy$$. Use this together with $x^2=t$ and a standard Gaussian integral $\endgroup$ – tired Sep 25 '15 at 12:43
  • $\begingroup$ @tired: perhaps you should offer this as an actual solution. $\endgroup$ – Ron Gordon Sep 25 '15 at 12:53
  • $\begingroup$ @tired I think this equality you hinted is worth the an explanation as an answer. If you could explain also for which kind of $f$ this equality holds it would be also very interesting. $\endgroup$ – Conrado Costa Sep 25 '15 at 18:39
  • $\begingroup$ @Conrado Costa maybe you want to have a look in the following paper Jack D'Aurizio posted a proof for the special case $f(x)=e^{-x}$. $\endgroup$ – tired Sep 26 '15 at 11:05
  • $\begingroup$ Here $\alpha$ is a positive real number, right? Do the formula holds also for $\alpha$ complex with Re$(\alpha)>0$, think the square root in the right hand side as the principal determintation of the square root? $\endgroup$ – foo90 Jun 5 '18 at 15:48

If $A,B>0$ $$\int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx =\sqrt{\frac{B}{A}}\int_{0}^{+\infty}\exp\left(-ABx^2-\frac{AB}{x^2}\right)\,dx\tag{1}$$ hence it is enough to compute: $$ I(C)=\int_{0}^{+\infty}\exp\left(-C^2 x^2-\frac{C^2}{x^2}\right)\,dx. \tag{2}$$ By splitting the integration range as $(0,1)\cup(1,+\infty)$ and setting $t+\frac{1}{t}=u$ we have: $$ I(C) = \int_{2}^{+\infty}\frac{t}{\sqrt{t^2-4}} e^{-C^2(t^2-2)}\,dt = \int_{0}^{+\infty}\frac{1}{\sqrt{u}}e^{-C^2(4u+2)}\,du\tag{3}$$ then setting $u=v^2$ we get: $$ I(C) = 2e^{-2C^2}\int_{0}^{+\infty}e^{-4C^2 v^2}\,dv = \color{red}{\frac{\sqrt{\pi}}{2C}\cdot e^{-2C^2}}.\tag{4}$$

  • 1
    $\begingroup$ Thanks Jack, that took me 3 hours to get it straight, but in the end every step is correct. I am amazed! $\endgroup$ – Conrado Costa Sep 25 '15 at 18:40

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